Almost-Sure Uniqueness and Versions

Why Uniqueness Is an Almost-Sure Statement

A conditional expectation is specified by measurability and integral identities. Those identities determine its values only up to sets of probability zero. Consequently, two functions can both satisfy the definition and disagree at some outcomes, while still representing the same conditional expectation.

This is not a defect in the definition. Integrals do not change when an integrand is altered on a null set. The important qualification is that a candidate must still be measurable with respect to the sigma-algebra on which we are conditioning. An arbitrary alteration on a null set need not preserve that measurability.

Definition and Terminology

Definition: Let \((\Omega,\mathcal F,\mathbb P)\) be a probability space, let \(\mathcal G\subseteq\mathcal F\) be a sigma-algebra, and let \(X\) be an integrable random variable. A random variable \(Y\) is a conditional expectation of \(X\) given \(\mathcal G\) if:

\(Y\) is \(\mathcal G\)-measurable and integrable; and
for every \(A\in\mathcal G\), \(\int_A Y\,d\mathbb P=\int_A X\,d\mathbb P\).

We write \(Y=\mathbb E[X\mid\mathcal G]\), with the understanding that this notation specifies \(Y\) only up to almost-sure equality.

Two random variables \(Y\) and \(Z\) are almost surely equal if \(\mathbb P(Y\ne Z)=0\). A version of a conditional expectation is a particular \(\mathcal G\)-measurable, integrable random variable satisfying the defining integral identities. Thus “the conditional expectation” can refer to the almost-sure equivalence class, while “a version” refers to one chosen representative.

A Basic Integral Fact

The uniqueness proof uses a simple consequence of nonnegativity. If a nonnegative integrable random variable has integral zero, it must vanish almost surely. We include the argument because it is the step that turns integral identities into an almost-sure conclusion.

Lemma: If \(H\ge 0\) almost surely, \(H\) is integrable, and \(\int_\Omega H\,d\mathbb P=0\), then \(H=0\) almost surely.

Proof. For each \(n\in\mathbb N\), the set \(A_n=\{H\ge 1/n\}\) is measurable. Nonnegativity gives \[ 0=\int_\Omega H\,d\mathbb P\ge \int_{A_n}H\,d\mathbb P\ge \frac{1}{n}\mathbb P(A_n). \] Therefore \(\mathbb P(A_n)=0\) for every \(n\). Since \(\{H>0\}=\bigcup_{n=1}^{\infty}A_n\), countable subadditivity implies \(\mathbb P(H>0)=0\). Hence \(H=0\) almost surely.

\square

Almost-Sure Uniqueness

Theorem: Suppose \(Y\) and \(Z\) are both conditional expectations of the same integrable random variable \(X\) given the same sigma-algebra \(\mathcal G\). Then \(Y=Z\) almost surely.

Proof. Both \(Y\) and \(Z\) are integrable and \(\mathcal G\)-measurable, so \(D=Y-Z\) is integrable and \(\mathcal G\)-measurable. For every \(A\in\mathcal G\), the defining identities give \[ \int_A D\,d\mathbb P = \int_A Y\,d\mathbb P-\int_A Z\,d\mathbb P = \int_A X\,d\mathbb P-\int_A X\,d\mathbb P =0. \] The sets \(A_+=\{D>0\}\) and \(A_-=\{D<0\}\) belong to \(\mathcal G\). On \(A_+\), \(D\) is nonnegative, and \[ \int_\Omega D\mathbf 1_{A_+}\,d\mathbb P =\int_{A_+}D\,d\mathbb P=0. \] The lemma applied to \(D\mathbf 1_{A_+}\) shows that \(D=0\) almost surely on \(A_+\). On \(A_-\), \(-D\) is nonnegative, and \[ \int_\Omega (-D)\mathbf 1_{A_-}\,d\mathbb P =-\int_{A_-}D\,d\mathbb P=0. \] The lemma applied to \((-D)\mathbf 1_{A_-}\) shows that \(D=0\) almost surely on \(A_-\) as well. Since \(\{D\ne0\}=A_+\cup A_-\), it follows that \(\mathbb P(Y\ne Z)=0\).

\square

The argument uses the defining identities only on two particular events, the positive and negative regions of the difference. Both events are available because the candidates are \(\mathcal G\)-measurable. The proof does not assert that \(Y(\omega)=Z(\omega)\) at every \(\omega\).

Almost-Surely Equal Candidates Are Both Versions

Proposition: Let \(Y\) be a conditional expectation of \(X\) given \(\mathcal G\). If \(Z\) is \(\mathcal G\)-measurable and \(Z=Y\) almost surely, then \(Z\) is also a conditional expectation of \(X\) given \(\mathcal G\).

Proof. Since \(Z=Y\) almost surely and \(Y\) is integrable, \(Z\) is integrable. For every \(A\in\mathcal G\), almost-sure equality implies equality of the integrals over \(A\): \[ \int_A Z\,d\mathbb P=\int_A Y\,d\mathbb P=\int_A X\,d\mathbb P. \] Together with the assumed \(\mathcal G\)-measurability, these are exactly the conditions in the definition.

\square

The measurability assumption on \(Z\) is essential. Almost-sure equality alone is not enough: the definition requires the chosen representative itself to be \(\mathcal G\)-measurable.

Worked Example: A Conditional Expectation with a Simple Conditioning Sigma-Algebra

Take \(\Omega=[0,1]\), let \(\mathcal F\) be the Borel sigma-algebra, let \(\mathbb P\) be Lebesgue probability measure, and let \(\mathcal G=\{\varnothing,\Omega\}\). Set \(X(t)=t\). The constant random variable \(Y(t)=1/2\) is \(\mathcal G\)-measurable and integrable. There are only two events to check. For the empty event both integrals are zero; for \(\Omega\), \[ \int_\Omega Y\,d\mathbb P=\frac12 \qquad\text{and}\qquad \int_\Omega X\,d\mathbb P=\int_0^1t\,dt=\frac12. \] Thus \(Y\) is a conditional expectation of \(X\) given \(\mathcal G\). In fact, every real-valued \(\mathcal G\)-measurable function is constant, and the identity on \(\Omega\) forces that constant to be \(1/2\). This illustrates uniqueness within the class of \(\mathcal G\)-measurable candidates.

Versions Can Differ Pointwise

Worked Example: Changing a Version on a Measurable Null Set

Again take \(\Omega=[0,1]\) with its Borel sigma-algebra and Lebesgue probability measure, but now let \(\mathcal G=\mathcal F\) and \(X(t)=t\). The random variable \(Y(t)=t\) is a conditional expectation of \(X\) given \(\mathcal G\), since it is \(X\) itself and therefore has the required integral over every Borel set.

Define \(Z(0)=7\) and \(Z(t)=t\) for \(t\ne0\). This is Borel-measurable: it differs from the Borel function \(t\mapsto t\) only on the Borel set \(\{0\}\). The two functions differ at \(0\), but \(\mathbb P(\{0\})=0\), so they agree almost surely. For each \(A\in\mathcal F\), their integrals over \(A\) are equal because the functions agree outside a null set. Hence \(Z\) is another version. The theorem says the versions agree almost surely, not everywhere.

Why Measurability Can Fail After a Modification

A null set for \(\mathbb P\) need not belong to the conditioning sigma-algebra. Therefore, changing a \(\mathcal G\)-measurable function on a null set can produce a function that is measurable with respect to \(\mathcal F\) but not with respect to \(\mathcal G\). The integral identities may survive the change, while the measurability requirement does not.

Worked Example: A Null Modification That Is Not a Version

Let \(\Omega=[0,1]\), let \(\mathcal F\) be the Borel sigma-algebra, let \(\mathbb P\) be Lebesgue probability measure, and take the trivial sigma-algebra \(\mathcal G=\{\varnothing,\Omega\}\). For \(X=0\), the constant function \(Y=0\) is a conditional expectation given \(\mathcal G\).

Now let \(W=\mathbf 1_{\{0\}}\). This is Borel-measurable and equals \(Y\) almost surely, because \(\mathbb P(\{0\})=0\). But \(W\) is not \(\mathcal G\)-measurable: the preimage \(W^{-1}(\{1\})=\{0\}\) is neither \(\varnothing\) nor \(\Omega\), so it does not belong to \(\mathcal G\). Thus \(W\) is not a conditional expectation given \(\mathcal G\), despite being almost surely equal to \(Y\). The condition that the candidate be \(\mathcal G\)-measurable cannot be omitted.

When Completeness Allows More Modifications

There is a precise condition under which an \(\mathcal F\)-measurable modification on a subset of a \(\mathcal G\)-measurable null event retains \(\mathcal G\)-measurability. Say that \(\mathcal G\) is complete relative to \(\mathcal F\) and \(\mathbb P\) if every \(\mathcal F\)-measurable subset of every \(\mathcal G\)-measurable null set belongs to \(\mathcal G\).

Proposition: Suppose \(\mathcal G\) is complete relative to \(\mathcal F\) and \(\mathbb P\). If \(Y\) is \(\mathcal G\)-measurable and \(Z\) is \(\mathcal F\)-measurable, and if \(Y=Z\) outside a set \(A\subseteq N\) where \(N\in\mathcal G\) and \(\mathbb P(N)=0\), then \(Z\) is \(\mathcal G\)-measurable.

Proof. For any Borel set \(B\subseteq\mathbb R\), the symmetric difference \(Y^{-1}(B)\mathbin{\triangle}Z^{-1}(B)\) is an \(\mathcal F\)-measurable subset of \(N\). By the stated completeness condition, this symmetric difference belongs to \(\mathcal G\). Also \(Y^{-1}(B)\in\mathcal G\). Since \[ Z^{-1}(B) = \bigl(Y^{-1}(B)\setminus (Y^{-1}(B)\mathbin{\triangle}Z^{-1}(B))\bigr) \cup \bigl(Z^{-1}(B)\setminus Y^{-1}(B)\bigr), \] and both pieces on the right belong to \(\mathcal G\), we have \(Z^{-1}(B)\in\mathcal G\). This holds for every Borel \(B\), so \(Z\) is \(\mathcal G\)-measurable.

\square

The qualification about \(N\) matters. Completeness relative to \(\mathcal F\) guarantees measurability for modifications on subsets of a \(\mathcal G\)-measurable null event. It does not, by itself, guarantee measurability for modifications on every \(\mathcal F\)-measurable null set.

For example, with the trivial \(\mathcal G=\{\varnothing,\Omega\}\) on Borel \([0,1]\), the completeness condition above holds vacuously: the only \(\mathcal G\)-measurable null event is \(\varnothing\). Yet the Borel null set \(\{0\}\) is not contained in any \(\mathcal G\)-measurable null event, and modifying the constant zero function at \(0\) produces the non-\(\mathcal G\)-measurable function from the previous example. To make a claim about every \(\mathcal F\)-measurable null set, one must instead assume that every such null set belongs to \(\mathcal G\).

What Uniqueness Does—and Does Not—Say

Almost-sure uniqueness gives a strong conclusion: once one conditional expectation exists, every other valid candidate agrees with it outside a \(\mathcal G\)-measurable null set. The disagreement set is \(\mathcal G\)-measurable because both candidates are \(\mathcal G\)-measurable.

It does not select a unique value at each individual outcome. Nor does it mean that every function obtained by changing values on a null set is a valid version. A useful way to keep the distinctions straight is:

Almost-sure equality concerns the probability of the disagreement set.
Versionhood also requires integrability and \(\mathcal G\)-measurability.
Completeness can ensure measurability for specified null-set modifications, but only under the stated containment and measurability conditions.

In applications, one may choose a convenient representative, but any pointwise statement about that representative should be interpreted with care. Statements invariant under changes on null sets are naturally statements about the almost-sure equivalence class.

Check Your Understanding

Try to answer each question using the definition and the distinction between a version and an almost-sure equivalence class.

Which two properties, in addition to the integral identities, must a candidate conditional expectation satisfy?
In the uniqueness proof, why do the sets \(\{Y-Z>0\}\) and \(\{Y-Z<0\}\) belong to the conditioning sigma-algebra?
Why does changing a version on a measurable null set preserve the integral identities?
Can an \(\mathcal F\)-measurable function that equals a conditional expectation almost surely fail to be another version? Explain.
What additional condition is needed to conclude that every \(\mathcal F\)-measurable null-set modification preserves \(\mathcal G\)-measurability?